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Source: MEMO 2016 I3

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#1 h Aug 24, 2016, 1:33 PM • 4 Y

Y by doxuanlong15052000, pisgood, Adventure10, Mango247

Let $ABC$ be an acute triangle such that $\angle BAC > 45^{\circ}$ with circumcenter $O$ . A point $P$ is chosen inside triangle $ABC$ such that $A, P, O, B$ are concyclic and the line $BP$ is perpendicular to the line $CP$ . A point $Q$ lies on the segment $BP$ such that the line $AQ$ is parallel to the line $PO$ .

Prove that $\angle QCB = \angle PCO$ .

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#2 h Aug 24, 2016, 7:36 PM • 4 Y

Y by pisgood, Tawan, Adventure10, Mango247

Lemma 1: Let $P, Q$ be points in the plane of $\triangle ABC$ and denote $R \equiv BP \cap CQ$ and $S \equiv BQ \cap CP.$ Let $\ell$ be the bisector of $\angle BAC.$ If $AP, AQ$ are symmetric in $\ell$ , then $AR, AS$ are symmetric in $\ell.$

Proof of Lemma: By the Dual of Desargues' Involution Theorem, there exists an involution that swaps $(AB, AC), (AP, AQ), (AR, AS).$ Since $AB, AC$ and $AP, AQ$ are symmetric in $\ell$ , this involution is reflection in $\ell.$ Hence $AR, AS$ are symmetric in $\ell.$ $\blacksquare$

Under inversion at $A$ , we obtain the following corollary.

Corollary: Let $P, Q$ be points in the plane of $\triangle ABC$ and denote by $R$ and $S$ the second intersections of $\odot(ABP), \odot(ACQ)$ and $\odot(ABQ), \odot(ACP).$ If $AP, AQ$ are symmetric in $\ell$ , then $AR, AS$ are symmetric in $\ell.$

Lemma 2: Let $P, Q$ be isogonal conjugates WRT $\triangle ABC.$ Then $\measuredangle BPC + \measuredangle BQC = \measuredangle BAC.$

Proof of Lemma: By examining $\triangle BQC$ , we have $\measuredangle BPC + \measuredangle BQC = \measuredangle BPC + \measuredangle QBC + \measuredangle BCQ = \measuredangle BPC + \measuredangle ABP + \measuredangle PCA = \measuredangle BAC,$ where the last step follows from examining quadrilateral $ABPC.$ $\blacksquare$
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Main Proof: Let $Q^*$ denote the second intersection of $\odot(BOC)$ and $\odot(AB)$ and let $R$ be the projection of $B$ onto $CA.$
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Since $BR, BO$ are symmetric in the bisector of $\angle ABC$ , the corollary for points $R, O$ and $\triangle ABC$ implies that $BP, BQ^*$ are symmetric in the bisector of $\angle ABC.$ Moreover, from $QA \parallel PO$ , we infer that $\measuredangle BQ^*A + \measuredangle BQA = 90^{\circ} + \measuredangle BPO = 90^{\circ} + \measuredangle BAO = \measuredangle BCA.$ Hence, by Lemma 2, $Q^*$ is uniquely determined as the isogonal conjugate of $Q$ WRT $\triangle ABC.$

Thus, using $\measuredangle BQ^*C = \measuredangle BOC = 2\measuredangle BAC$ , we obtain $\measuredangle BQC = -\measuredangle BAC$ by Lemma 2. $(\star)$

By $(\star)$ we have $\measuredangle PCQ = \measuredangle BQC - 90^{\circ} = \measuredangle OCB.$ Hence, $\measuredangle PCO = \measuredangle QCB$ , as desired.

This post has been edited 1 time. Last edited by Dukejukem, Aug 24, 2016, 11:05 PM
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#3 h Aug 25, 2016, 8:29 PM • 4 Y

Y by pisgood, ILIILIIILIIIIL, Adventure10, Mango247

Let $\measuredangle ABP=\beta$ and $\measuredangle QAB=\theta$ $\Longrightarrow$ $\measuredangle PAQ$ $=$ $\beta+\theta$ , so from $PO\parallel AQ$ and $ABPO$ is cyclic we get $\measuredangle PAQ$ $=$ $\measuredangle PQA$ $=$ $\measuredangle OAB$ $=$ $\measuredangle OBA$ $=$ $\beta+\theta...(\star)$ . Let $R\equiv BP\cap \odot (ABC)$ $\Longrightarrow$ $\measuredangle ARB$ $=$ $\measuredangle ACB$ $=90^{\circ}$ $-\beta$ $-\theta$ , so from $\measuredangle AQB$ $=$ $\beta$ $+$ $\theta$ we get $\measuredangle QAR$ $=$ $90^{\circ}$ , but by $(\star)$ we get $PA$ $=$ $PQ$ , since $\measuredangle QAR$ $=$ $90^{\circ}$ and $PA$ $=$ $PQ$ we get $PA$ $=$ $PQ$ $=$ $PR$ , since $CP\perp QR$ we get $CQ$ $=$ $CR$ $\Longrightarrow$ $\measuredangle BAC$ $=$ $\measuredangle CRQ$ $=$ $\measuredangle CQR...(\star\star)$ . Since $O$ is the circumcenter of $\odot (ABC)$ we get $\measuredangle OCB$ $=$ $90^{\circ}-\measuredangle BAC$ . On the other hand, so from $(\star\star)$ we get $\measuredangle PCQ$ $=$ $90^{\circ}-\measuredangle PQC$ $=$ $90^{\circ}-\measuredangle BAC$ hence $\measuredangle OCB$ $=$ $\measuredangle PCQ$ $\Longrightarrow$ $\measuredangle PCO=\measuredangle QCB.$
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This post has been edited 1 time. Last edited by FabrizioFelen, Aug 27, 2016, 5:14 AM
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#4 h Aug 27, 2016, 2:47 AM • 2 Y

Y by pisgood, Adventure10

From the conditions, $A,P,O,B$ lie on a circle in this order ( $\alpha>45^\circ$ ).

Need: $\angle PCQ=\angle BCO=90^\circ-\alpha$ , same as $\frac{CP}{QP}=\tan \alpha$ . $(1)$

$M$ and $N$ : centers of $BPC$ and $APOB$ , resp.
$MN$ is perpendicular to their radical axis, $BP$ .
Also $OM\perp BC$ and $ON\perp AB$ , so $\angle ONM=\angle ABP=\varphi$ , $\angle OMN=\angle CBP=\theta$ .
From $\triangle OMN$ , we have $\frac{OM}{ON}=\frac{\sin \varphi}{\sin \theta}$ . $(2)$

Let $\rho$ be the radius of circle $APOB$ and $R$ be the circumradius.
$\rho=\frac{AB}{2\sin AOB}=R\cdot \frac1{2\cos \gamma}$ .
Since $\angle PQA=\angle QPO=\angle BPO=\angle BAO=90^\circ-\gamma$ , and $\angle BPA=2\gamma$ , we get $QP=AP$ , so
$QP=AP=2\rho \sin \varphi=R\frac1{\cos\gamma}\sin\varphi$ $(3)$
but also
$CP=BC\sin \theta=2R\sin \alpha\sin \theta$ , $(4)$
$\frac{OM}{ON}=\frac{R\cos\alpha}{\rho}=2\cos\alpha\cos\gamma$ . $(5)$

Now $(2-5)$ imply $(1)$ , so done.

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#5 h Aug 29, 2016, 2:43 AM • 4 Y

Y by bgn, K.titu, Adventure10, Mango247

Finally got around to solving this problem.

Angle chasing gives $\angle AQP = 90 - \angle C$ and $\angle APB = \angle AOB = 2 \angle C$ .
This gives $PA=PQ$ . Now if we set $R = BP \cap \omega$ ( $\omega$ is the circumcircle of $\triangle ABC$ ) we have $\angle ARB = \angle C$ , so $\triangle QAR$ is a right triangle. This gives $PQ=PR$ , and it is easy to see that $CP \perp RQ$ , so $\angle PCQ = 90 - \angle PRC = 90-\angle A = \angle OCB$ , which gives the desired.

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#6 h May 4, 2019, 1:37 PM • 2 Y

Y by Adventure10, Mango247

A bit difficult.

Let $M$ be the midpoint of $BC$ and let $O'$ be the circumcenter of $\triangle BQC$ . Let $S$ and $T$ be the circumcenters of $\triangle APB, \triangle AQB$ respectively. As
$\angle AQB = 180^{\circ}-\angle BPO = 90^{\circ} + \angle C$ , we get that $\angle TAB =\angle C$ or $S$ is midpoint of $OT$ . But $MS$ and $O'T$ are both perpendicular to $CP$ , thus $M$ is the midpoint of $OO'$ .

Hence $\angle BQC = 180^{\circ}-\angle A$ so $\angle PCQ=90^{\circ}-\angle A$ , implying the desired isogonality.

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#7 h Mar 17, 2024, 1:20 PM

Y by

$\textbf{Lemma}$ Define the projections of $P$ on $AB, AC$ as $X, Y$ and as $Z, W$ the projections of $Q$ . Then $P, Q$ are isogonal conjugates wrt $\angle BAC$ if and only if $X, Y, Z, W$ are concyclic. Moreover the midpoint of $PQ$ is the center of this circle.

The proof is a simple angle chasing exercise. The problem statement is equivalent with the isogonal conjugacy of $Q, O$ wrt $\angle PCB$ . Let $M$ be the midpoint of $BC$ and $D$ be another intersection of $AQ$ with the circumcircle of $A, P, O, B$ . Then $PQDO$ is a parallelogram. Denote $N$ as the intersection of its diagonals. By our lemma the statement is equivalent with showing that $ND = NP = NM$ which is equivalent to $\angle PMD = 90^{\circ}$ . However, $PM= MC= MB$ and $PD = OB$ since $PODB$ is an isosceles trapezoid. By this observation also $\angle DPM = \angle OBM$ , thus $\triangle OBM \equiv \triangle DPM$ , hence $\angle PMD = \angle OMB = 90^{\circ }$ .

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#8 h Mar 18, 2024, 11:02 PM

Y by

Elegant conjuring problem! Posting for storage.

danepale wrote:

Let $ABC$ be an acute triangle such that $\angle BAC > 45^{\circ}$ with circumcenter $O$ . A point $P$ is chosen inside triangle $ABC$ such that $A, P, O, B$ are concyclic and the line $BP$ is perpendicular to the line $CP$ . A point $Q$ lies on the segment $BP$ such that the line $AQ$ is parallel to the line $PO$ .

Prove that $\angle QCB = \angle PCO$ .

Observe that $\angle QAO=\angle POA = \angle PBA$ and $\angle PAO=\angle PBO$ hence $\angle PAQ = \angle OBA$ . Likewise, $\angle PQA = \angle QAB+\angle QBA=\angle OAB-\angle OAQ+\angle PBA=\angle OAB$ since $\angle OAQ=\angle PBA$ . Thus, $PA=PQ$ , and so if we reflect $Q$ across $P$ , we get point $R$ such that $\angle RAQ=90^{\circ}$ , and $\angle ARB=\angle ARQ = 90^{\circ}-\angle AQP=90^{\circ}-\angle OAB=\angle C$ , hence $R$ lies on the circumcircle of $ABC$ . Since $CP \perp QR$ , it follows that $\angle CQB=180^{\circ}-\angle CQP=180^{\circ}-\angle CRP=180^{\circ}-\angle CAB$ hence $\angle QCP=(180^{\circ}-\angle CAB)-90^{\circ} = 90^{\circ}-\angle CAB=\angle OCB$ so $CO, CQ$ are isogonal in angle $PCB$ , which proves our claim.

This post has been edited 1 time. Last edited by anantmudgal09, Mar 18, 2024, 11:02 PM

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